IMC 2006 · Day 2 · P10

By passing to a subspace we can assume that $v_1, \dots, v_n$ are linearly independent over the reals. Then there exist $\lambda_1, \dots, \lambda_n \in \mathbb{R}$ satisfying $$v_{n+1} = \sum_{j=1}^{n} \lambda_j v_j$$ We shall prove that $\lambda_j$ is rational for all $j$. From $$-2 \langle v_i, v_j \rangle = |v_i - v_j|^2 - |v_i|^2 - |v_j|^2$$ we get that $\langle v_i, v_j \rangle$ is rational for all $i, j$. Define $A$ to be the rational $n \times n$-matrix $A_{ij} = \langle v_i, v_j \rangle$, $w \in \mathbb{Q}^n$ to be the vector $w_i = \langle v_i, v_{n+1} \rangle$, and $\lambda \in \mathbb{R}^n$ to be the vector $(\lambda_i)_i$. Then, $$\langle v_i, v_{n+1} \rangle = \sum_{j=1}^{n} \lambda_j \langle v_i, v_j \rangle$$ gives $A \lambda = w$. Since $v_1, \dots, v_n$ are linearly independent, $A$ is invertible. The entries of $A^{-1}$ are rationals, therefore $\lambda = A^{-1} w \in \mathbb{Q}^n$, and we are done.